A101880 Number of arrangements of the partitions of n (e.g., 111 counts for 6).
1, 1, 3, 9, 35, 161, 913, 6103, 47319, 416235, 4092155, 44424095, 527511445, 6798907249, 94504286703, 1408973416617, 22426222745159, 379522092608177, 6804315177704869, 128828842646944135, 2568533750228603835, 53788282243854336411, 1180357162840624656959
Offset: 0
Examples
a(3) = 9 as we have 3, 12 (2) and 111 (6). a(4) = 35 as 4, 31 (2), 22 (2), 211 (6) and 1111 (24).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..450
- Jon Perry, Partition Tables [broken link]
Programs
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Maple
b:= proc(n, i, p) option remember; `if`(n=0, p!, `if`(i<1, 0, add(b(n-i*j, i-1, p+j), j=0..n/i))) end: a:= n-> b(n$2, 0): seq(a(n), n=0..25); # Alois P. Heinz, Apr 06 2016 -
Mathematica
Rest[ CoefficientList[ Series[ Sum[ n!x^n / Product[1 - x^k, {k, n}], {n, 20}], {x, 0, 20}], x]] (* Robert G. Wilson v, Feb 10 2005 *) -
Sage
from sage.combinat.partition import number_of_partitions_length def A101880(n): return sum(number_of_partitions_length(n, k)*factorial(k) for k in (0..n)) print([A101880(n) for n in (0..21)]) # Peter Luschny, Aug 01 2015
Formula
a(n) = Sum_{i=1..n} P(n,i)*i!, where P(n,i) is the number of partitions of n into i parts.
G.f.: Sum_{n=1..infinity} (n!*x^n / Product_{k=1..n} (1-x^k)). - Vladeta Jovovic, Jan 29 2005
G.f.: ( 1 - G(0) )/(1-x) where G(k) = 1 - (k+1)/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 22 2013
a(n) ~ n! * (1 + 1/n + 2/n^2 + 5/n^3 + 16/n^4 + 60/n^5 + 253/n^6 + 1180/n^7 + 6023/n^8 + 33306/n^9 + 197719/n^10 + ...), for coefficients see A331826. - Vaclav Kotesovec, Jan 28 2020
Extensions
More terms from Vladeta Jovovic, Jan 29 2005
a(0)=1 prepended by Alois P. Heinz, Apr 06 2016
Comments